The man lands 12.1 m far from the base of the cliff, so he will survive
Explanation:
The motion of the person is a projectile motion, which consists of two independent motions: Â
- A uniform motion (constant velocity) along the horizontal direction Â
- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction Â
We start by considering the vertical motion, to find the time of flight of the person, with the following suvat equation:
[tex]s=ut+\frac{1}{2}gt^2[/tex]
where :
s = 50 m is the vertical displacement of the man (the height of the cliff)
t is the time of flight
u = 0 is the initial vertical velocity
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
Solving for t, Â we find
[tex]t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(50)}{9.8}}=3.19 s[/tex]
Now we know that the man moves horizontally, with a constant velocity of
[tex]v_x = 3.8 m/s[/tex]
Therefore, the horizontal distance covered by the man during this time is
[tex]d=v_x t = (3.8)(3.19)=12.1 m[/tex]
So the man lands 12.1 m far from the base of the cliff: therefore, on the lake, so he will survive.
Learn more about projectile motion:
brainly.com/question/8751410
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